# Thread: Find All Values of "A" That Create No Solution for "X"

1. ## Find All Values of "A" That Create No Solution for "X"

Hello everybody!

These two problems have been boggling me for awhile now:

#1)1/(1+1/x) = a Find all values of "a" that yield no solution for "x"

#2)(6x-a)/(x-3) = 3 Find all values of "a" that yield no solution for "x"

I'm unsure of how to solve for this type of solution, but I've tried isolating the "x" variables like so:

#1) 1/(1+1/x) = a

1) 1 = a + a/x

2) 1 - a = a/x

3) x/a = 1/(1-a)

4) x = a/(1-a)

#2) (6x-a)/(x-3) = 3

1) 6x - a = 3x - 9

2)
3x - a = -9

3) 3x = a - 9

4) x = a/3 - 3

However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!

2. ## Re: Find All Values of "A" That Create No Solution for "X"

Originally Posted by Lexielai
Hello everybody!

These two problems have been boggling me for awhile now:

#1)1/(1+1/x) = a Find all values of "a" that yield no solution for "x"

#2)(6x-a)/(x-3) = 3 Find all values of "a" that yield no solution for "x"

I'm unsure of how to solve for this type of solution, but I've tried isolating the "x" variables like so:

#1) 1/(1+1/x) = a

1) 1 = a + a/x

2) 1 - a = a/x

3) x/a = 1/(1-a)

4) x = a/(1-a)

#2) (6x-a)/(x-3) = 3

1) 6x - a = 3x - 9

2)
3x - a = -9

3) 3x = a - 9

4) x = a/3 - 3

However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
You almost have it for the first problem. What values of a produce no solution for x? (Take a look at what a can't be.)

For the second one, I see no reason to restrict the value of a. Did you type it correctly?

-Dan

3. ## Re: Find All Values of "A" That Create No Solution for "X"

I see what you mean with the first problem. It's easy to discern logically. I'm still confused with the second one though, and I did type it exactly like the original problem.

My friend told me the answer was "18" after plugging in values, but I'm more concerned with a proper method to find the answer than the answer itself. Any help would be appreciated!

4. ## Re: Find All Values of "A" That Create No Solution for "X"

Originally Posted by Lexielai
Hello everybody!

These two problems have been boggling me for awhile now:

#1)1/(1+1/x) = a Find all values of "a" that yield no solution for "x"

#2)(6x-a)/(x-3) = 3 Find all values of "a" that yield no solution for "x"

I'm unsure of how to solve for this type of solution, but I've tried isolating the "x" variables like so:

#1) 1/(1+1/x) = a

1) 1 = a + a/x

2) 1 - a = a/x

3) x/a = 1/(1-a)

4) x = a/(1-a)

#2) (6x-a)/(x-3) = 3

1) 6x - a = 3x - 9

2)
3x - a = -9

3) 3x = a - 9

4) x = a/3 - 3

However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.

The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$

Now solve for x. $x = \dfrac{a}{1 - a} \implies a \ne 1.$ And $0 \ne x = \dfrac{a}{1 - a} \implies a \ne 0.$

So I would answer problem 1 as a is not equal to either 0 or 1.

In the second problem, it is clear that $x \ne 3.$

Now solving for x we get $\dfrac{6x - a}{x - 3} = 3 \implies 6x - a = 3x - 9 \implies 3x = a - 9 \implies x = \dfrac{a - 9}{3}.$

$3 \ne x = \dfrac{a - 9}{3} \implies 9 \ne a - 9 \implies a \ne 18.$

Solving for a we get $a = 3x + 9,$ which raises no problems. So a is not equal to 18.

Does this help

5. ## Re: Find All Values of "A" That Create No Solution for "X"

Originally Posted by JeffM
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.

The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$

Now solve for x. $x = \dfrac{a}{1 - a} \implies a \ne 1.$ And $0 \ne x = \dfrac{a}{1 - a} \implies a \ne 0.$

So I would answer problem 1 as a is not equal to either 0 or 1.

In the second problem, it is clear that $x \ne 3.$

Now solving for x we get $\dfrac{6x - a}{x - 3} = 3 \implies 6x - a = 3x - 9 \implies 3x = a - 9 \implies x = \dfrac{a - 9}{3}.$

$3 \ne x = \dfrac{a - 9}{3} \implies 9 \ne a - 9 \implies a \ne 18.$

Solving for a we get $a = 3x + 9,$ which raises no problems. So a is not equal to 18.

Does this help
I don't agree with this.

Let a=18

$\dfrac {6x-18}{x-3}=\dfrac {6(x-3)}{x-3}=6$

There's nothing invalid about x=3 if a=18.

6. ## Re: Find All Values of "A" That Create No Solution for "X"

Hi,
If you know about functions, domain and range, maybe the following will help.

1. Rephrase the problem as f(x)=1/(1+1/x); the domain of f is then all reals except x=0 and x=-1. Now what value a is not in the range of f? Answer a=0 and a=1.

2. Again rephrase the problem. f(x)=(6x-a)/(x-3) for some a. For any a, the domain of f is all reals except x=3. Now the horizontal asymptote is y=6, no matter the value of a unequal to 18. (If a=18, the graph of f is the horizontal line y=6 with a hole at x=3) So the range of f is all real numbers y except y=6 except when a=18. So the only value of a so that f(x)=3 can not be solved for x is a=18. You might want to graph f for various a values.

7. ## Re: Find All Values of "A" That Create No Solution for "X"

Originally Posted by romsek
I don't agree with this.

Let a=18

$\dfrac {6x-18}{x-3}=\dfrac {6(x-3)}{x-3}=6$

There's nothing invalid about x=3 if a=18.
Yes, there is. $\displaystyle \frac{6x- 18}{x- 3}= 6$ ONLY if x is NOT 3.

$\displaystyle f(x)= \frac{6x- 18}{x- 3}= 6\frac{x- 3}{x- 3}$ has domain "all real numbers except 3.
The graph of y= f(x) is the horizontal line y= 6 with a hole at x= 3.

8. ## Re: Find All Values of "A" That Create No Solution for "X"

Originally Posted by HallsofIvy
Yes, there is. $\displaystyle \frac{6x- 18}{x- 3}= 6$ ONLY if x is NOT 3.

$\displaystyle f(x)= \frac{6x- 18}{x- 3}= 6\frac{x- 3}{x- 3}$ has domain "all real numbers except 3.
The graph of y= f(x) is the horizontal line y= 6 with a hole at x= 3.
Really? This is the first time I've seen it stated that you can't always divide out common factors.

Not the first time I've been wrong.

9. ## Re: Find All Values of "A" That Create No Solution for "X"

Saying that $\displaystyle \frac{6x- 18}{x- 3}= 6\frac{x- 1}{x- 1}= 6$ when x= 1 is the same as saying that $\displaystyle \frac{0}{0}= 1$. It isn't.
$\displaystyle f(x)= \frac{6x- 18}{x- 3}$ and $\displaystyle g(x)= 6$ are almost the same but are not exactly the same.

10. ## Re: Find All Values of "A" That Create No Solution for "X"

Originally Posted by romsek
Really? This is the first time I've seen it stated that you can't always divide out common factors.

Not the first time I've been wrong.
Like anyone that has studied Calculus for too long you might be trying to recall
$\displaystyle \lim_{x \to 3} \frac{6(x - 3)}{x - 3} = 6$

which is perfectly valid.

-Dan